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NSWChemistryQuick questions
Module 6: Acid/Base Reactions
Quick questions on Enthalpy of neutralisation and calorimetry explained: HSC Chemistry Module 6
10short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.
What is why weak combinations release less heat?Show answer
If either the acid or the base (or both) is weak, two things happen in sequence:
What is calorimetric procedure?Show answer
A polystyrene cup (or insulated foam cup) is an excellent simple calorimeter because polystyrene has very low thermal conductivity.
What is sources of error?Show answer
To reduce these errors, use a stirred, insulated calorimeter, a graphed temperature-time correction, and equal initial temperatures for the two reactants.
What is step 4: Comparison?Show answer
The strong-strong value is $-57.6$ kJ/mol. The weak-acid case is about 2 kJ/mol less exothermic because the ionisation of $CH_3COOH$ absorbs roughly that amount, consistent with the trend.
What is forgetting the negative sign?Show answer
$\Delta H$ for an exothermic reaction is negative. The solution gains heat (positive $q$ for the solution), so the reaction releases heat ($\Delta H < 0$).
What is using only one volume in $m$?Show answer
The mass in $q = mc\Delta T$ is the total mass of the combined solution, not just the acid or the base.
What is mixing up moles of acid and moles of water?Show answer
When a diprotic acid like $H_2SO_4$ neutralises a monoprotic base, each mole of acid produces 2 mol of water. Always express $\Delta H$ per mole of water formed.
What is quoting $-57.6$ kJ/mol for a weak combination?Show answer
Weak combinations release less heat. Use the experimental value, not the textbook strong-strong value.
What is calling heat loss the only error?Show answer
Heat loss is the dominant systematic error, but markers reward students who also mention the cup absorbing heat, $c$ approximation, and slow-mixing effects.
What is using degrees C in $\Delta T$ vs K?Show answer
A temperature change in degrees C equals the same change in K. Either unit is fine in $\Delta T$; just keep $c$ in matching units (J/g/K).